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A 1100 kg car rolling on a horizontal surface has speed v = 70 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

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    The car comes to rest (temporarily)at compression distance X, at which point the initial kiteci energy (1/2) M V^2 is converted to spring potential energt, (1/2)kX^2.

    Therefore
    kX^2 = mV^2
    k = m (V/X)^2

    Make sure you convert V to meters per second before using the formula. k will be in Newtons/meter

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